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Electric fields

  1. Jul 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Two charges each +4 microcoloumbs ar eon the x axis, one at the origin and one at x=8m. Find the electic field at the x axis at x=10m and x=2m.

    2. Relevant equations

    E=kq/r

    3. The attempt at a solution

    I tried the x=10m first. What I did was take E1=kq/r^2, or (8.99x10^9*4mC)/8^2 and E2=kq/r^2, or (8.99 x 10^9*4mC)/2^2, and then added E1 and E2 up.

    And it was wrong. Am I doing it wrong?
     
  2. jcsd
  3. Jul 8, 2007 #2

    Dick

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    For one thing the distances from x=10 of x=0 and x=8 are 2 and 10. Not 2 and 8.
     
    Last edited: Jul 8, 2007
  4. Jul 8, 2007 #3
    Oh really? What I did was say that since x=10, one of them would be 2m and the other would be 8m, like 10-2=8, right? Like it would be 8m b/t the two charges and then 2m b/t charge 2 and the point.
     
  5. Jul 8, 2007 #4

    Dick

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    You are measuring the distance from x=10 where you want to find the E field. On source is 2m away and the other is 10m away. The distance between the two charges is not important.
     
  6. Jul 8, 2007 #5
    Oh oh oh...okay, I was just about to ask about that.

    So I'm to use 2 and 8 for the second one (when x=2m), right? Because one charge is 2m away and the other is 8m away this time?

    Problem is, since the location is in the middle this time, and because it's two positive charges, I think it's going to be zero in the middle (since the charges will repel and cancel out).
     
  7. Jul 8, 2007 #6

    Dick

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    The two charges will only cancel out if the position is midway between the charges (since they are equal). I don't think x=2m is midway between 0m and 8m. Just do it the same way you did the first problem. Except consider the direction of the two forces and don't just blindly add.
     
  8. Jul 8, 2007 #7
    So one's going to the positive x direction, and the other's going to the negative x direction...

    Oh, okay okay...I've come to both correct answers now. Thanks.
     
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